By using properties of definite integrals, evaluate the integrals ∫π2−π2sin7xdx.
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Solution
∫π2−π2sin7xdx. Here, f(x)=sin7xf(−x)=sin7(−x)=[−sinx]7=−sin7x∴f(−x)=−f(x)So, f(x) is an odd function.∫π2−π2sin7xdx.=0[∵∫a−af(x)dx=0if f(x) is an odd function